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Rosita is writing an explicit function for the geometric sequence:

80,40,20,10,dots
She comes up with

t(n)=160((1)/(2))^(n)". "
What domain should Rosita use for 
t so it generates the sequence?
Choose 1 answer:
(A) 
n >= 0 where 
n is an integer
(B) 
n >= 0 where 
n is any number
(c) 
n >= 1 where 
n is an integer
(D) 
n >= 1 where 
n is any number

Rosita is writing an explicit function for the geometric sequence:\newline80,40,20,10, 80,40,20,10, \ldots \newlineShe comes up with\newlinet(n)=160(12)n t(n)=160\left(\frac{1}{2}\right)^{n} \text {. } \newlineWhat domain should Rosita use for t t so it generates the sequence?\newlineChoose 11 answer:\newline(A) n0 n \geq 0 where n n is an integer\newline(B) n0 n \geq 0 where n n is any number\newline(C) n1 n \geq 1 where n n is an integer\newline(D) n1 n \geq 1 where n n is any number

Full solution

Q. Rosita is writing an explicit function for the geometric sequence:\newline80,40,20,10, 80,40,20,10, \ldots \newlineShe comes up with\newlinet(n)=160(12)n t(n)=160\left(\frac{1}{2}\right)^{n} \text {. } \newlineWhat domain should Rosita use for t t so it generates the sequence?\newlineChoose 11 answer:\newline(A) n0 n \geq 0 where n n is an integer\newline(B) n0 n \geq 0 where n n is any number\newline(C) n1 n \geq 1 where n n is an integer\newline(D) n1 n \geq 1 where n n is any number
  1. Step 11: Plugging in n=1 n = 1 : Rosita's function for the geometric sequence is t(n)=160×(12)n t(n) = 160 \times \left(\frac{1}{2}\right)^n . To determine the domain, we need to see which values of n n will generate the sequence 80,40,20,10, 80, 40, 20, 10, and so on. Let's start by plugging in n=1 n = 1 to see if we get the first term of the sequence, which is 80 80 .
    t(1)=160×(12)1 t(1) = 160 \times \left(\frac{1}{2}\right)^1
    t(1)=160×12 t(1) = 160 \times \frac{1}{2}
    t(1)=80 t(1) = 80
  2. Step 22: Checking for n=2n = 2: Since t(1)=80t(1) = 80, which is the first term of the sequence, we can see that the sequence starts with n=1n = 1. Now let's check if the function works for n=2n = 2 to get the second term of the sequence, which is 4040.\newlinet(2)=160×(12)2t(2) = 160 \times (\frac{1}{2})^2\newlinet(2)=160×14t(2) = 160 \times \frac{1}{4}\newlinet(2)=40t(2) = 40
  3. Step 33: Testing non-integer value: The function also works for n=2 n = 2 , giving us the second term of the sequence. This suggests that the domain should start at n=1 n = 1 . Now let's check if the function only works for integer values of n n by trying a non-integer value, such as n=1.5 n = 1.5 .
    t(1.5)=160×(12)1.5 t(1.5) = 160 \times \left(\frac{1}{2}\right)^{1.5}
    t(1.5)=160×(12) t(1.5) = 160 \times \left(\frac{1}{\sqrt{2}}\right)
    t(1.5)=1602 t(1.5) = \frac{160}{\sqrt{2}}
    This result is not an integer and does not correspond to any term in the geometric sequence.
  4. Step 44: Determining the domain: Since the function does not produce terms of the geometric sequence for non-integer values of nn, we can conclude that the domain should only include integer values of nn. Therefore, the domain should be n1n \geq 1 where nn is an integer.

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